text:
Design glider without aileron
What difficulties would it take to build a glider without ailerons?  I read that "if you can control the turns with the rudder maybe you don't need ailerons."  Well, apart from the fact that it would only slide to one side, I was wondering if there was a rudder-elevator combination that would allow bank angle turns or is it something that only ailerons can do.  (We must be able to achieve with 42° bank angle)

Is it limited to using the rudder only?
I think you could achieve a fair deal of roll control with asymmetrical spoiler deflection.
But this would be problematic for low-speed control (as they also destroy a lot of lift)

There is also roll due to instantaneous yaw, see this link

Flex-wing hang gliders have no ailerons-- google "Wills Wing" or "Moyes".  Many rigid-wing hg's have spoilers rather than ailerons-- google "ATOS".  Many radio-control model gliders use only rudder and elevators for control, relying on lots of dihedral for yaw-roll (or more properly slip-roll) coupling to achieve a roll torque and bank the wing.  Google "Gentle Lady" or "Radian RC glider". The earlier versions of the "Super Floater" ultralight sailplane did the same -- you can google that too but it will take some doing, as most links go to later versions which had ailerons, and less dihedral.

Answers are going to be a bit vague and subjective the way the question is currently phrased.  Because you are asking about 3 things:  Design, build, and fly.  (I know, I'm being nit picky...)  The first two are simple, just don't design or build it with ailerons.  BUT, if you want it to fly well, (specifying 42 degrees desired AOB is a useful performance parameter) you may need to account for the omission.  More details would be helpful, like size, speed, etc.  (many RC planes don't have ailerons)

If ones does a really good job matching the wing and tail, the wing will pull the glider through the turn once bank is established, and the tail will provide sufficient directional stability (weathervane) to keep the nose into the wind.  But, sometimes, deliberately flying uncoordinated (forward slip) is a very useful tool to control glide angle.  Ailerons as needed for this.

Without ailerons you can get a bank angle only as a side effect of the sideslip introduced by the rudder.
This side effect mainly depends on the sweep and dihedral angles and on the position of the wing (high, middle, low).
So if your bank angle has to be perfectly controllable (as I suppose it should be) then it would be better to rely on ailerons.
Sideslip is also caused by lateral gusts, so if you design an airplane which rolls only due to sideslip then it is going to be sensitive also to lateral gusts.

By setting the rudder up very high on a high tail one could increase its bank effect. You would have to counter steer: turning the rudder to the right would wrangle the plane into a left bank.
In effect this is a vertical aileron; imagine a second rudder underneath the plane that turns the opposite direction.

Yes, in the roll axis ailerons can work vertically.  Some planes even split the horizontal stabilizer (stabilator) up and down.  But "wrangling" the plane increases the drag.

One really would not want to design a glider without ailerons because:

minimizing drag is key to achieving the greatest lift to drag ratio.

You glide much farther by flying coordinated.  With ailerons, you set your bank.  The rudder need only keep the nose "into the wind", compensating for adverse yaw from the downward aileron.
Without ailerons, one must push the nose out of the relative wind with the rudder in order to create a "crosswind component", which rolls the aircraft.  This does work, but is far more draggy.
But spoilerons are an interesting alternative, currently being used on many long range aircraft where fuel economy is of primary importance.
One may test to see if the rudder/vertical stabilizer, usually shaped as a vertical airfoil, is actually of lower drag (in combination with ailerons) than raising a spoiler into the airstream in order to turn.


text:
Use twist to improve stall characteristics in ideal distribution of lift coefficient
Is practical use twist (geometric or multiple airfoils to improve stall characteristics) when it has lift coefficient distribution ideal or near due to wing plataform?
I am analyzing a "Schuemann" wing, and I am in the dilemma of whether to use a twist (multiple airfoils or geometrical) to improve stall and look for compromise or investigate other forms directly (vortex generator, e.g.).

What do you mean with improve stall characteristics? An higher max AoA? A milder stall? An earlier stall warming?

Stall warning. I chose the airfoil with higher max AOA and docile stall. Thanks

@andy  What type of ship is this...a sailplane?? powered aircraft?  Is this full scale, or RC?

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

sort: votes


text:
Incidence angle in Sailplane
I am determining the shape of the fuselage of a sailplane and I found something that gives me a little conflict, and that is that in most sailplanes it has an inclined tip, I read that this helps reduce drag, but I also see that the wing is not at some notable angle of incidence for the best L/D, with this in mind three questions arose:

Do sailplane wings adjust just right for best L/D? because I also had the doubt that adjusting it at this point causes any inclination to drop this relationship and I don't know if it would be optimal to use degrees before the best L/D.
Does the tilt of the nose affect this angle or is it only designed for drag? I also read about visibility.
Is figure a and b equivalent? not in terms of drag, but in a possible configuration without considering pilot visibility.

Where did you read that it reduces drag?

Does "inclined tip" mean washout?

What do you mean by "the wings adjust?"  What does the adjusting - the designer, the ailerons, the aeroelasticity?

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Cruise-climb in glider with electric motor?
I have been looking for the definition of cruise-climb for gliders that use an electric motor, but so far I have only been able to find a definition based on liquid fuel.  I'm asking because that condition appears in an RFP for a self-propelled glider with an electric motor.
"An aeroplane cruising technique resulting in a net increase in altitude as the aeroplane mass decrease"

Clearly, that RFP isn't using the definition you quoted for that term.  (Either that, or the author didn't consider what he's writing about very carefully.) Can you quote a little more of the RFP, or link to it, to provide more context?

I share it https://www.google.com/url?sa=t&source=web&rct=j&opi=89978449&url=https://www.aiaa.org/docs/default-source/uploadedfiles/membership-and-communities/university-students/design-competitions/graduate-team-aircraft-design-competition/2023-24-aiaa_sl_trainer_rfp_r2.pdf%3Fsfvrsn%3D75e1ff52_0&ved=2ahUKEwjqwpiIt6mCAxVFSzABHbzzA0sQFnoECBIQAQ&usg=AOvVaw2W-jHVdyp866i8z940SQfp

andy please note that the answers of Robert and Rob, although being formally correct, do not really apply to the RFP you are analysing. The RFP itself contains all the info that you need and you should only refer to that, see for example the comments of @quietflyer

"Cruise climb" does have a definition regarding a cruising aircraft rising to a higher altitude as it loses weight from fuel consumption.  This relates more towards long range aircraft such as airliners.
A second application of "cruise climb", is one of climbing at an airspeed greater than Vy.
Looking at it in the scope of Vx, Vy, and cruise climb, we see excess thrust and Angle of Attack can be used for a steeper climb, a faster climb, or greater distance while climbing, respectively.

because angle of attack is lower at higher airspeeds, cruise climbing may be a more efficient way of gaining distance while climbing.

Why cruise climb with a glider?  It wouldn't matter if it was gas or electric.  One needs to go and find another thermal.
Cruise climbing the glider is really an extension of "cleaning up" slats and flaps, minimizing drag, and gaining as much distance as possible for a given amount of energy.

Reading the RFP one can easily see what is meant with "cruise-climb" (requirement 1.a.ii):

The self-retrieve begins with a self-launch and includes 30 minutes of soaring flight before the glider is unsuccessful in finding atmospheric lift to continue soaring flight and must use motor power to “self-retrieve”. This includes a climb, cruise-climb, then descent during which the aircraft must cover a distance of at least 50 km, ending at an altitude at least 1000 feet above the airport.

So the cruise-climb relates to the need for the glider to cover 50 km and reach an altitude of 1000 feet if it hasn't encounter any good thermal in the previous half an hour, as simple as that.

How is "cruise climb" different from the "climb" segment, in the RFP's meaning?

@RalphJ -- see the tables in the RFP -- for the “Design reference Instructional Flight" mission, for the "climb" segment, a climb rate of at least 600 fpm is specified.  So that it doesn't forever to get to a reasonable altitude to shut down and start looking for thermals.   Also the target shutdown altitude is 2500' above airport altitude.  For the "Design reference self-retrieve flight" mission, it is a little more complicated.  Two different "climbs" are mentioned, the first being identical to the one mentioned above, and also a "cruise climb", where no minimum vert speed is specified.

(Distance covered in "cruise climb" counts toward a target 50 km range for self-retrieve.  A portion of that 50 km self-retrieve range appears to be intended to be accomplished in a power-off glide.)

Constant mass battery electric aircraft have no need to cruise climb.
Cruise climb is a technique where you allow the aircraft to slowly climb during cruise to keep the specific range at its optimal value.  You're typically gaining altitude to maintain constant lift coefficient at constant Mach number (or true airspeed, above the tropopause these are the same).
A constant mass aircraft has no need to cruise climb.  Just cruise.

I woudl rather say that constant mass battery electric acft are unable to cruise climb. They are destined to fly the whole time at constant mass and that is one of their disadvantages.

@Jpe61 Not losing weight in cruise is a small disadvantage, but cruise climb is a maneuver generally not allowed by ATC and seldom taken true advantage of.  Long range flights can at best hope to step climb.  Battery energy density is enough of a disadvantage that BE aircraft don't really have to complain about being constant mass.

My observation (as a passenger) does not match that. On almost all flights I can think of there certainly has been multiple upwards altitude changes enroute.

Check out FlightAware or similar flight tracking websites.  A quick random sample seems to show that transcontinental (USA) flights at night (low traffic time) sometimes show a nice step-climb, typically three segments across the continent.  Daytime flights show more erratic altitude changes, likely due to traffic and weather, often descending mid-flight.  Shorter flights (half that distance) seem to generally hold constant altitude.  I'm sure trans-oceanic (long and lower traffic) get to step climb better -- I have not seen any flight tracks with a true cruise climb.

I'll have inter- and transcontinental samples to report soon  but you are right in that true cruise climb is rare. Intermittent step ups are common in my experience.


text:
Is part 141 ground school really supposed to be 35hours?
I'm in a part 141 program and in ground school. My instructor dismisses class very early almost every day, but in the logbook he keeps for us he writes a lot more time than we actually meet on any given topic.  Is that okay?

A Ground School Syllabus needs to be checked and approved by the FAA.  As part of that approval, they will decide if it meets minimum requirements, and is expected to take the required hours.
If the actual delivery of the material in ground school takes a bit less (or more) time than approved, that is OK, so long as the approved syllabus is closely followed.
Source: https://www.faa.gov/documentLibrary/media/Advisory_Circular/AC_141-1B.pdf


text:
How does the engine wash improve engine performance?
According to maintenance manual, we usually carry out engine wash (CFM56) with pure water as scheduled task. Pure water is injected into the engine from the LP compressor inlet while the engine is cranked.
I wonder if the engine performance would improve significantly after having a wash.
What is cleaned from the engine gas path? is it carbon residue or something else.
As it is said that the water should be injected at 1 o'clock position on the LPC (aft view), why does it need to be at that specific position? does it clean certain components at that position?
Thanks for your explanation.

That reminds me my glider trainer that told me washing the wings helps not loosing one point of L/D ratio

A performance increase after a compressor wash is expected to happen, but it does not return to the original performance as there are more degradation mechanisms at play. This answer explains that organic build-up causes a change in airfoil shape causing the recoverable performance degradation. Another source may be deposition of salts and dissolved constituents to deposit on the blades as a result of ingesting salt water from which the water evaporates in the hotter stages of the compressor. E.g. erosion is a non-recoverable degradation (for a compressor wash). This is clearly shown in the figure below.

From: Modern Gas Turbine Systems, High Efficiency, Low Emission, Fuel Flexible Power Generation, Woodhead Publishing Series in Energy

I wonder if the engine performance would improve significantly after having a wash.

The performance increase is considered significant enough, else we would not go through all the trouble of performing the procedure. Fuel costs is one of the major direct operating costs, reducing fuel costs by washing the engine doesn't only save money, it also reduces the amount of used fuel (nice green image for the airline).

As it is said that the water should be injected at 1 o'clock position on the LPC (aft view), why does it need to be at that specific position?

This could be done to prevent build-up of water in the lower section; now it has to travel half the engine down, by the time it reaches that location the water will be transported aft by the churning blades. The water is therefore better distributed, which is confirmed by this practical answer.

Could you please include what exactly the performance factor is? I'm thinking that the engine thrust should still be somewhat similar if not identical even if the engines are dirty because of the engine control computers, or at least on engines that are EPR controlled. If the EPR and thus the thrust are not as they should be the FADEC would just inject more fuel. So engine thrust would be the same just at a cost of more fuel burn. Is this what the performance factor describes?

@Jan Frequently the exhaust gas temperature is a direct measure. Engines are EPR or N (fan spool speed) controlled. Indeed, fouling means that more fuel needs to be added, to maintain thrust, but that will make the engine deteriorate faster, so you want to clean it once in a while to prevent accelerated damage accumulation.

The compressor blades accumulate a coating of organic material from the atmosphere on the blades, which in the later stages, where the temperatures get above the material's flash point, may be mostly carbon from cooked organic material like pollen and bugs (the same sort of blackened accumulated crud you'll see on the butterfly of a late stage compressor bleed shutoff valve when you remove it).  It reduces the efficiency of the compressor since the blade contours are altered in subtle ways by the crud coating.  The water bath is just to erode that stuff off.  It's basically the same thing as cleaning bugs and dirt off wing leading edges.
Not sure about the reason for the clocking instruction.  Either to clean something like a probe at that position, or conversely, (and more likely by my guess) to avoid excessive water impingement on a probe or sensor on the other side.

I work within a engine test facility. During a compressor wash process we inject water and detergent. This also depends on how much work have been carried out during the overhaul on the engine.
The positions are given only to provide an even flow of water through the engine.
Sadly I have no experience with live aircraft, so I can not comment when their on wing.

I wonder how long it takes to perform engine wash and what crank the engine in test facility. As we use pneumatic starter, which only allowed to run in few minutes, therefore, we cannot wash for too long.

Hi. Same in our test facility, we turn the engine pneumatically so we still have to adhere to starter limits. Each wash cycle take 3-5 minute depending on engine types.

Thanks. I also want to know if only compressor would benefit from engine wash. How about combustion chamber and turbine? do we necessarily perform seperate turbine or combustion chamber wash?

Yes it would benefits the combustion chamber and turbine, once the VBV’s are scheduled closed the water will penetrate the whole core. Core wash at the test facility take around 2 hours to complete. Do you schedule VBV’s when completing a core wash on the wing? How to you motor the engine? Thanks

Yes, we do close the VBV when washing engine. We use its starter which pneumatically powered by bleed air from Apu. It also takes around 2 hours to wash, however, mostly for preparation and close-up. The main process takes only 30mins each engine. Cheers

Thanks. So your process is quicker then ours. Is there any noise regulations at the airport? How often is it required to wash an engine? Because I work for GE 90% of the engines we wash are GE90-115b.

Yes, we have noise regulation, therefore, we are allowed to crank engine or have engine run at idle. In case of running at high power, we have specific area where engine can be run at high power. It is scheduled by engineering department, I dont know if it is period or because the engine performance is out of limit or other reasons. BTW, we work with A320 equipped with CFM 56-5B and PW1000. Cheers


text:
Has anyone been able to verify ADS-B works without radar?
I know that one of the biggest advantages of ADS-B is "supposed" to be that it will directly communicate between equipped aircraft without ground support. Has anyone actually seen that in action? That is, seeing another aircraft on an ADS-B display where radar coverage is impossible, say, a remote valley surrounded by mountains with no radar station near.
Obviously both aircraft would need to be ADS-B equipped for this to work.

Yes, I've seen traffic in areas without radar coverage.

I don't understand what you mean by both aircraft have to be ADS-B equipped. One ADS-B OUT equipped aircraft and an ADS-B receiver is enough to verify that ADS-B works

Use dump1090 with a SDR dongle, this is what all the crowdsourced flight trackers use (Flightaware, Flightradar24, etc)

I can confirm that ADS-B works without radar. I have been involved in several ADS-B trials in areas where there was no or only partial radar coverage. It works. No surprise, it has been designed that way.

The purpose of ADS-B In is to detect and display the ADS-B Out aircraft and related information. Radar is not a component of ADS-B; so by definition it can operate in the absence of radar.  However, since not all aircraft are or will be equipped with ADS-B Out, in the US, a service called Traffic Information Service-Broadcast (TIS-B) is provided. TIS-B depends upon the presence of radar and proximity to an ADS-B (ground) radio station. A transponder only aircraft (no ADS-B Out) is detected by radar.  The track information on that aircraft is broadcast to properly equipped ADS-B Out & In aircraft in that proximity. So in a remote valley (or any area with no radar), the ADS-B In aircraft would have to capable of detecting an ADS-B Out aircraft directly on the correct link (1090Mhz or 978Mhz).

I agree with your technical explaination, but you cannot really say that radar is not a part of it. There are two different frequencies and two different systems (as you said), mode-ES at 1090mhz and UAT at 978Mhz. Instead of unifying on one system and frequency, the FAA decided that they would serve as the "translator" between the systems, that is, when the radar site receives a GPS position transmission on one system, it rebroadcasts it on the other system.

(continued) This is a stunning bad idea, since it not only creates a situation where large and small aircraft cannot see each other directly, but even in radar coverage, it introduces a delay in the reception of the signals between aircraft. ADS-B normally has the interesting property that the signal is stronger and more accurate as aircraft approach each other, because GPS is more accurate if in near locations vs. far. The translation requirement abrogates that.

The translator function is provided by ADS-R (Rebroadcast).  It receives the broadcast on one frequency, translates it and broadcasts that information on the other frequency.  This functionality resides in a control station for multiple ADS-B ground radios. It does not use any radar input. The delay it introduces in undetectable to a pilot. The accuracy of GPS reported position does not depend on relative proximity of the aircraft to each other.

By "radar" I was referring to the radar installation. I know what primary and secondary radar is.  Sorry, it just isn't true that the "accuracy of GPS reported position does not depend on relative proximity of the aircraft to each other", otherwise WAAS would have no reason to work. GPS has path accuracy issues, which is why WAAS works. It gives you a series of error references so that you get the net error from a station near to you, and can subtract the local accuracy out. Two GPS receivers will give a more similar answer depending on how close they are to each other.

You also neatly side stepped the other issue. I KNOW what a rebroadcaster is. It adds both round trip delay, and several miles of round trip delay (i.e., the path delay times 2) is not NOTHING. Next, it adds another point of failure that didn't exist before, that is, the operation of a ground station, which was not even supposed to be necessary in the original definition of ADS-B. Further, the ground station can be working perfectly, but simply be below reception range, meaning that you and the heavy about to run you over cannot see each other.

The "solution" to this FAA ABORTION of technology has come from the ADS-B equipment makers, which is so called "dual band" technology, meaning each receiver is able to see both 1090 and 978 frequencies (and protocols). It works, they have made it relatively cheap, but that does not excuse what the FAA did, which is try to please everyone, and end up screwing everyone.

I am sorry to be testy here, but I have followed ADS-B throughout it's development, had these arguments with several of the proponents, including the head of the AOPA, and I get very tired of people talking to me like I am a poor idiot pilot needing to be educated. The FAA screwed the pooch on this design. I notice that even the AOPA later said it was a screwed up design.

By the way, if you are convinced rebroadcasting is such a good idea, I suggest you have a talk with Garmin, Lynx, and virtually all of the makers of ADS-B dual band receivers. They know quite well that waiting for a ground station to rebroadcast the signal from one band to another is a bad idea. The "theory" of 1090 rebroadcast was that heavies would always be in radar contact since they fly high and descend into radar service areas. The problem is, there are plenty of relatively large passenger aircraft, like 50-100 seat that serve small airports and might not always be in radar areas.

The so called "regional" carriers must have reacted in horror to the idea that they were dependent on ground repeaters to see small airplanes headed to the same airports they were. So now were are we? We have two different systems, 1090 and UAT, two different protocols, and twice the equipment expense. The rebroadcast is a joke. ADS-B makers are shipping dual band receivers as fast as they can make them.

Indeed.  Thank you for making clear your position.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Why does Thrust Reverser need to be operative during the cargo door operation?
It is stated that when the Electric Pump push button is depressed, the electric pump is automatically running as long as the cargo door selector valve is set to either OPEN or CLOSE position.
During this period, the operation of other systems powered by the Yellow Hydraulic system is inhibited except Alternate Brake and Thrust Reverser of ENG #2.

The Alternate Braking is understandably essential as it stops aircraft from moving, and cargo door operation is considered that aircraft is in parking, not landing phase when the thrust reversers are only being used.

My question here is why thrust reverser ENG#2 needs to be operative during cargo door operation.
Thanks for your answers.

(Source: Training Manual A320 Family)

It doesn't need to be operative, it just isn't inoperative.

My guess from looking at distribution diagrams is that there simply isn't a valve in between the alternate braking and engine 2 thrust reverser that could close, but I'm not sure why that is the case?

“It is stated that when the Electric Pump push button is depressed, the electric pump is automatically running as long as the cargo door selector valve is set to either OPEN or CLOSE position”
This is not correct. Cargo door operation will automatically energize the Elec Pump while it is in the OFF position. (not depressed)
“My question here is why thrust reverser ENG#2 needs to be operative during cargo door operation”
It doesn’t need to be operative. It needs to be Not Inhibited,  in case there is a malfunction.
The cargo door uses the Yellow system. It is designed to be used on the ground even with the Electric Pump in the OFF position. The pump will come on automatically in this case to supply hydraulic power to the cargo door. It will inhibit all the  other items on the yellow system to insure adequate hydraulic power to the cargo door.
BUT......If there was a malfunction in the Cargo Door system while in flight, it could possibly cause the Yellow Elec Pump to engergize while it is in the OFF position. If that were to happen, it is designed to inhibit all items on the Yellow system. That would be bad, so they designed the switch to inhibit everything except Alt Brakes and #2 Reverse.
This situation could only happen if the Yellow Elec Pump was automatically commanded ON while in the OFF position.


text:
Why is engine #1 still used when taxiing with one engine?
According to the answer in this post Taxiing with one engine: Is engine #1 always used or do they switch? Engine #1 is used because it pressurizes the green HYD system which powers the Nose Wheel Steering. I found that, in latest model of A320 and A321, NWS is driven by yellow HYD system that is pressurised by engine #2. However, pilots still use only engine #1 if they want to taxi with one engine running.
Is there other reason for using engine #1?

  Source: Airbus AMM


text:
Why does A320 Main Wheel need to be tightened 3 times during installation?
I have another question related to maintenance practices. During replacing MLG wheel of A320, it is said that we need to tighten the wheel nut 3 times with different torque values.
The first one has the highest value. Then, we need to loosen and tighten the wheel nut again, however, with lower torque values.
In my opinion, I think that the highest torque value is to put the wheel all the way on the axle, however, I was unable to find the answer for why we need to torque the nut again for 2 different values.
And are there any other reasons for tightening the wheel nut with different value?
Thanks for your answer.

That is intriguing! My automotive maintenance experience is that sometimes you torque everything to a lower setting (ensuring the part is mounted evenly), then torque to the final, higher setting. Additionally, some bolts are "stretch bolts" that are one-time use - the incredibly high torque setting literally stretches the bolt and that is what holds it from coming lose. Tightening to a high setting then loosening and re-torquing to a lower setting is very counter intuitive!

I also have the same question. According to the AMM, we need to torque the wheel nut initially with the highest torque (for example 240lb.ft) . Then loosen it, tighten it again with second value (for example 70 lb.ft), and torque with final value (for example 110 lb.ft)

FreeMan - Commercial jet aircraft wheels are installed with a single nut on the axle, like a wheel hub on a car.
The commercial jet aircraft axle is a few inches in diameter and is hollow. The wheel bearings are tapered, and while the axle nut is being tightened the wheel is spun ih the normal direction of rotation. A higher initial torque is applied to the nut, and the pressure on the bearings helps to get the races properly seated while the bearings are rolling over them.
The nut is loosened, then the process is repeated using the torque specs specified in the manual.
One of the airlines I worked for had a policy that required that the final time the nut was tightened, it would be turned by hand. After a while, they went back to following the manufacturer's recommendations.
To recap, really tight to seat the bearings, looser for in service operation.
I hope that helps.

Welcome to Aviation.SE!


text:
Why would there be airflow from the back of a TAT probe?
During walk-around checks, I occasionally find airflow coming out from the back of the TAT probes on the Airbus A320. I wonder if it's air to cool down the TATs as they are heated during flight for anti-ice purposes.
I had a look at the description and operation of the TAT probes in the AMM but I couldn't find any information about that airflow and where it comes from.

This is a so called aspirated TAT sensor, optional on the Airbus. Air is being drawn through aspirated TAT sensors to ensure that they correctly indicate the temperature even when the aircraft is not moving.
Normal TAT sensors develop a bias when the aircraft is static.
This video gives a nice explanation.

Thanks DeltaLima, very clear explanation.


text:
Why can Boeing V-22 Osprey carry more cargo internally than externally?
According to the Wikipedia, V-22 Osprey can lift 20,000lb of internal cargo, whereas it is able to carry only 15,000lb externally by dual hook.
I wonder why there is a difference between external and internal cargo capacity.
Is it due to the aerodynamics effect of the cargo?

May also be due to structural issues, load carrying structures of the hooks may not be able to take more.

Is a load connected to the hooks a "slung load"? Maybe the pendulum effect of a hanging load makes it harder to control at the max weight.

Hovering is the most inefficient, most power-hungry mode of flight. Is it possible that the 20000 pounds are for a rolling start with horizontal velocity using partially tilted rotors?

The 20000lb of internal cargo is not going to be a point load, whereas the hook is connected to 1 (or is it 2) point(s) - so it's like having 7.5 or 15,000lb of load in those two spaces [a much strong frame is required for that load than the spread out internal load].

The difference is because when the cargo is inside the aircraft the weight is spread across the cargo compartment floor, whereas when it is carried externally, the weight is concentrated entirely on one or both cargo hooks. The cargo hook attachment points are reinforced, but they still have a weight limit. When the cargo is inside the aircraft the weight limit is the MTOW of the aircraft.

Like the first answer, but with more detail. The rating on the cargo hook attachment points is going to be less than the entire aircraft & it always will be. If you increase the rating of the attachment points you have to up the overall aircraft rating.

The only way to increase the rating of the attachment points is to increase their number to spread out the weight.

"If you increase the rating of the attachment points you have to up the overall aircraft rating." Why?  I see no fundamental reason why even a single attachment point couldn't be designed/built to be stronger than the aircraft can lift, and indeed it seems sensible, to avoid failure due to shock-loading, and to allow for potential future up-rating of the engines

I may have said that un-clearly. If the aircraft can carry  20,000 lbs over the floor of the aircraft to carry the same amount on one point would require that one attachment point to carry 20K. It would be massively over-engineered to carry that weight on different points of the aircraft. You are correct, over-engineering it might make sense if you know you're going to massively increase engine power.


text:
Could a pilot navigate using distances and radials from an RNAV waypoint, similar to VOR navigation?
I know that typically RNAV navigation involves a list of waypoints, and the aircraft simply flies from one point to the next.  What I'm curious about is if a procedure could be developed (as a special engine out procedure, for example) that would be something like

Depart on runway heading. At distance 2nm from FOBAR waypoint, turn
  direct to FOBAR. Track radial 090 outbound from FOBAR.

Would the instruments in an RNAV equipped aircraft allow a procedure like that to be flown?  Would it require a certain system (say only from a certain manufacturer), or require additional systems (like an FMS)?
CLARIFICATION FROM COMMENTS:
I understand that a procedure like the one above would not be typical or preferred, but in very hazardous terrain (for example) with no real other options, would the pilot simply have the tools available to be able to fly it?
If it matters, I'm particularly interested in commercial aviation and answers that would be applicable to any airport worldwide.

IFR engine-out is an emergency, and ATC will make sure to clear you direct wherever you want to go. I'm not sure why you would be departing with an engine out, IFR or not, unless you had a death wish.

@RonBeyer I know of many airports at the bottom of a narrow valley, surrounded by large cliffs on all sides, where the safest way out could be following a procedure similar to the one I mention.

Even then, you say as a special "engine out" procedure, why would you depart an airport in the mountains with an engine out? If you are engine out coming in that may be different, however ATC will give you vectors, not waypoints.

@RonBeyer: Anytime you depart anywhere, there is a chance of it becoming an engine out departure, and that possibility needs to be accounted for.

That may be true, however trying to do some kind of VOR-ish RNAV procedure with an already high workload is going to be a recipe for disaster. That is why ATC will just issue vectors/altitudes. In an emergency ATC's job is to make yours as a pilot easier. If I were flying engine out and declared on departure and heard that from the controller, provided I make it back alive I'd walk up to that controller and slap them.

Oh I fully understand it's not the preferred option.  I just wanted to know if it was possible.  Wouldn't the terrain in an area like that prevent ATC from being able to track you, especially at low altitudes just after takeoff?

@RonBeyer Actually, it is the operators responsibility to develop emergency turn procedures where they are needed. They are typically referenced to navaids/RNAV-fixes. (Sometimes as simple as "Passing 1.5 DME, turn right TRK 120", but sometimes more complex, involving multiple navaids and possibly climb in hold). ATC can assist you with vectoring above their MVA, but the emergency turn procedure is required to get to MVA/MSA safely. At those low heights, you might not even be visible on radar.

You might want to remove your 'engine out' point completely, it's irrelevant to your basic question and seems to be a distraction, judging by the comments.

@Pondlife: I added that part hoping to avoid "why would you even want to try to do that" responses.  It seems I failed miserably =(.

@GeneralMike, could you clarify the avionics fit and/or the exact difficulty faced, when you say " with no real other options, would the pilot simply have the tools available to be able to fly it?"

@RonBeyer, what part of this question suggests the OP is considering initiating takeoff with an engine out?  (You mentioned it twice above…)  Engines can fail after V1, and that’s why you have procedures to deal with them.

In a modern B737 (NG/MAX) (and probably most other comparable aircraft), this is trivial. The simplest (and most realistic option for an emergency turn procedure) would be to fly the procedure in HDG SEL:

Enter FOBAR on FIX page
Add a 2 nm range ring for FOBAR on the FIX page. This ring will be displayed on the ND (Navigation Display)
Add the 90 radial for FOBAR on the FIX page. This radial will be displayed on the ND.
Fly the procedure in HDG SEL, monitoring the ND.

I figured it would probably be possible with modern aircraft, thanks for confirming.  But what if you only had older equipment, say one of the earliest RNAV systems and no FMS?

Technically, as long as you can define custom waypoints, either by lat/long or radial/distance from existing fix, sure. Is it something that you would see in practical use? Not so sure.

One of the aircraft that I fly has dual Avidyne IFD540 GPS units onboard. I can select any waypoint that is in my database and treat the waypoint as though it were a VOR in-so-much-as I can set the GPS in OBS mode and then manually select any "radial" to track to or from that waypoint no different than if it were an actual VFR.
I believe that the Garmin units can do the same thing.
Once I have this dialed in, the autopilot will follow the "radial" that I have selected no different than a VOR.
In the situation you are describing I think if they were going to be tracking outbound from FOBAR you would be going "somewhere" and instead of tracking outbound they would simply be going to their next waypoint.
Ultimately, while not something that one would normally do, with modern navigation equipment, it can be done.

Not only is this possible, but some RNAV procedures (such as the JFK DEEZZ5) require it.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Can TODA ever be shorter than TORA, under any circumstance?
I've always been told that TODA* = TORA† + Clearway.  I can't really imagine what a negative clearway would be, so that leads me to think that it is required that TODA >= TORA.  Interestingly, a colleague pointed out a certain airport (Invercargill, New Zealand - ICAO code NZNV) where the declared distances published by a certain source describe a TODA that is in fact less than the TORA.
My colleague and I are under the impression that some kind of typographical error has occurred when this source was publishing the data.  However, I can't help but think back to strikingly similar conversations we had after seeing a previously unheard of case where ASDA‡ < TORA.  As it turned out, the FAA had changed how Runway Safety Areas were considered, in such a way that it is totally possible for ASDA < TORA (and we've been seeing it more and more frequently ever since).
*TODA = Take-Off Distance Available†TORA = Take-Off Run Available‡ASDA = Accelerate-Stop Distance Available

TL;DR
I wanted to see if anyone has ever heard of any other case where TODA < TORA, under any regulatory agency.  What would the rationale be if that were the case?

If runway has obstacle at the end maybe then TODA < TORA? Taking opposite way TODA will be > TORA.

Can you be less mysterious about the "certain" airport and sources? A specific example could be very helpful here.

Since different Civil Aviation Authorities could, at least in theory, use different definitions of TODA/TORA/etc., it would probably be good to phrase this question in the context of just one country's regulations, at least if the title question is what you're wanting to know.

@reirab: Good point.  I guess I'm curious if anyone is aware of any CAA anywhere that would allow for this.  Updated.

No
Takeoff Distance Available (TODA) can never be shorter than Takeoff Run Available (TORA).
AIM 4-3-6

Takeoff Distance Available (TODA) – The takeoff run available plus the length of any remaining runway or clearway beyond the far end of the takeoff run available.

I had previously downloaded the FAA runway database and determined there is no TODA less than TORA.  The mysterious airport may have incorrect data or the database is incorrect.
FAA Airport Information
The same database shows that approximately 14% of the runways have an ASDA shorter than TORA.  Multi-engine pilots need to be aware of how their aircrafts performance was created and how to apply the correct declared distances.

I've updated my OP - the airport was Invercargill, New Zealand - ICAO code NZNV.  I wonder if being an international airport makes any difference.

@GeneralMike Well, it does at least mean that the AIM and FARs are irrelevant, but NZ's CAA appears to use the same definition of TODA, according to Part 1 of their Civil Aviation Rules (PDF).

Looking up declared distances from APG show these values for runway 4 at NZNV:  TORA 7087 ft, TODA 7284 ft, ASDA 7087, LDA 6660.  Runway 22 values at NZNV: TORA TODA ASDA all 7251 ft, LDA 6693 ft.  Their AIP is not very helpful.

What your question is essentially asking is, aside from the theoretical possibility that some nation uses TORA and TODA, but defines them differently from every other country, is there an airfield where there is an area designated as a take-off surface that is suitable for the take-off ground-run of an aircraft, but not suitable to fly over. That is clearly not possible.


text:
Why do I keep getting different results to an aircraft's turn radius when I plug variables into different calculators and solve on my own?
I'm trying to solve for an aircraft's turn radius at mach 3.17 (2432 mph) and a bank angle of 19.8; however, after using various online calculators and personally solving the equation, I keep getting different answers.
I'm using the aircraft turn radius equation:
R = V^2/(11.26 ⋅ tan(θ))
R = Aircraft Turn Radius, V = Nm/h,θ = Bank Angle
Plugging in the variables of mach 3.17 and a 19.8 degree bank angle yields the following
R = 2096^2/(11.26 ⋅ tan(19.8))
(mach 3.17 ≈ 2096 nm/h)
R = 4393216/(11.26 ⋅ 0.36002)
R = 4393216/ 4.05384
R = 1,083,717 ft
But after plugging the same variables into various online calculators and getting different results, I'm lost. What did I do wrong?

Can you link to the calculators that give different results? And show what you entered there? Your calculation seems to be correct. This calculator also gives a very similar result with your numbers.

Hey, thanks for the comment! This is one of the calculators I used, but after plugging in the airspeed (mach 3.17) and bank angle (19.8) it gives a turn radius of "330038.4579509235" feet. Did I just plug these variables into the calculator incorrectly?

This is how it looks to me. Make sure to select the correct units and it gives ~1,083,715ft.

You say "this", but don't provide a link, or even a name. Makes it hard for others to find. ;)

The calculator hyperlinked in the first comment is one of the calculators I used that gave me a different result.

Units are the root of all evil in most of these situations.
The original equation has a constant of 11.26. This would correspond to a Velocity expressed in some particular unit (knots, mph, km/h, m/s, furlongs/fortnight, fraction of C, whatever) and the output radius in some unit (nm, sm, km, inches, mm, leagues, AU, whatever).
Unless you know that the units of your input match those that the constant applies to, and are understanding the formula's output in the correct units, then it's a case of garbage in - garbage out.
That applies both to your equation and to the online calculators. If a calculator expects its input in mph and delivers its output in yards, but you give it an input in knots & believe the output to be in meters, you'll have a moderately close but definitely incorrect answer.
Unfortunately, not all sources of formulas are as clear about such things as they could be, which is why (since omnipresent calculators render arithmetic errors mostly obsolete) inconsistent units are the most common source of problems such as this.
One related thought: using Mach # for velocity is problematic, because the speed of sound is not a single constant. It varies with temperature, so an aircraft at Mach X.xx at 10,000' has a different speed (expressed as ft/miles/meters/furlongs per second/minute/hour/fortnight) than an aircraft flying at the same Mach number at 30,000' where the temperature is much colder. Entering every calculator with a speed not expressed in Mach will alleviate this issue. Conversions between knots, mph, km/h, m/s, and so on are exact; conversion between any of those and Mach isn't (unless temperature is specified).

Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on [meta], or in [chat]. Comments continuing discussion may be removed.

The above answer is fairly accurate but it doesn't address the question. If you're plugging the same numbers into different calculators and getting different answers, the problem is the machine you're using. What is the decimal precision of the device you're using?
The calculator cannot represents every possible number accurately so errors are introduced. That's why when we're expecting to see 1, 0.9999784 sometimes shows up in the display. It becomes very glaring when you're dividing big numbers by small ones. There's a whole field of mathematics that solves this problem, numerical analysis.

What the questioner is talking about is very big differences, not machine-epsilon differences


text:
What is the maximum speed at which thrust vectoring can be used? Thrust Vectoring on the SR-71
The SR-71 is limited to a 45 degree bank turn at speeds in between mach 2.85 to ~3.2. This means the blackbird would take almost 5 1/2 minutes to make a 180 degree turn at mach 3. Would it be possible to add thrust vectoring to increase maneuverability and decrease such an exaggerated turn radius?
What is the maximum speed an aircraft has used thrust vectoring?

The problem is not the speed-per-se, but the G forces induced by the resulting maneuver. Given the same control surface deflection angle, the higher the speed the higher the G force exerted on the aircraft (and the pilot). In the case of the SR-71 the issue was also the small wing-area and as a result the low maximum allowed wing-loading, as answered on https://aviation.stackexchange.com/questions/8028/what-is-the-minimum-turning-radius-of-an-sr-71

@DeepSpace in part, disruption of air flow into the engines at high bank angles and aoa is also a factor.

Maybe not what you're after, but most launch vehicles use thrust vectoring as their primary control.  It is also used by on-orbit spacecraft -- not always a gimbaled nozzle, but sometimes that way and other times as discrete thrusters.
I am sure some tactical missiles  would meet your criteria of high speed uses of thrust vectoring.

Usually known as engine gimbaling rather than thrust vectoring for launchers. STS gimbaling test.

Would it be possible to add thrust vectoring to decrease turn radius?

It depends.
Vectoring thrust does help in maximising the turn rate but according to which kind of turn is flown.

In a instantaneous turn, i.e. a turn where we don't care about losing speed and/or altitude, then vectoring thrust maximises the turn rate if thrust becomes perpendicular to the flight path. Hawker Siddeley Harrier can employ this technique.

In a sustained turn, i.e. a turn flown keeping speed and altitude (which I suppose is what you are considering), then vectoring thrust maximises the turn rate if thrust is aligned with the flight path. In a sustained turn as the one we are talking about with a so big turn radius, thrust is basically already aligned with the flight path and vectoring it wouldn't bring any noticeable improvement.

Obviously all this is true as long as the load factor and the stall constraints are respected. Generally speaking turn rate $\dot{\psi}$ equals:
$\dot{\psi}=\frac{g\sqrt{n-1}}{V}$
so that it increases with higher Gs (higher load factor $n$) but decreases with velocity $V$.

When you use the variable Velocity (V), what unit is it in? Mph?

@SamJones: everything is in SI system, i.e. velocity in m/s, g in m/s² and psi in rad/s


text:
Can composite material be used in leading edge of airfoil instead of aluminium through Additive manufacturing?
I want to know about, is it possible to replace aluminium by additively manufactured leading edge of airfoil by changing different infill patterns to improve the impct strength from bird impact on it.

The title and the subject seem to be asking slightly different questions here.  What do you mean by "composite material"? Carbon fiber composite?  What do you mean by "additively manufactured"? Laser powder bed fusion?  Are you proposing to replace an traditional aluminum lead edge with an AM leading edge?  or replace an aluminum leading edge with a carbon fiber composite edge?

Lots of aircraft use composite material for their leading edges.
There is no need to use additive manufacturing to accomplish this.
That said, you might be interested in filament winding, automated tape laying, and automated fiber placement machines.


text:
Why do you land a fighter jet with AOA Units?
I currently begann learning Falcon BMS and watched some fighter cockpit landings on YT and learned the technique to land a fighter is to use your AOA and not Airspeed like you would do it on a GA Plane or on an Airliner what is the reason for that ? why are you not calculating the Approach Speeds like you would do it on an Airliner with X-Wind and weights etc. ?

Because AOA is the metric that governs all airfoil aerodynamics. And airline pilots also use AOA, indirectly. They determine (calculate) their landing speeds based on their configuration, (flap/slats settings), and gross weight and this is, effectively, calculating what speed to fly to get that same AOA, (that is the highest AOA, sufficiently below the stall AOA to allow safe stall margin). This allows landing at the slowest speed possible without risk of stalling.
The reason this is done for fighters and large aircraft, and not so much for GA aircraft is because the speed that gets you to that best AOA changes due to gross weight, and landing gross weights, (as a percentage of total gross weight), vary significantly more for fighters and large aircraft than they vary for most GA aircraft. So the landing speed for a small cessna does not need to be adjusted so much for the weight of the plane.
In the F-4, for example, the manual had **two ways to determine how to fly on final approach.

Calculate and fly by speed:
a. Full Flaps: 140 Kias + 2 kts / 1000 lbs of fuel.
b. Half flaps: 145 Kias + 2 kts / 1000 lbs of fuel.
c. No flaps:   150 Kias + 2 kts / 1000 lbs of fuel.

Or...

Just fly "On-Speed" AOA, referencing one of the several AOA Indications.

**Note: numbers in option 1 are from memory... They not be 100% accurate!

"+ 2 kts / 1000 lbs of fuel" -- Does that mean +2 kts per 1000 lbs of fuel?

Yes. That's exactly what it means.

Angle of attack is the most convenient metric of the performance of the wing. One can safely assume the wing will for example always stall at the same AoA regardless of weight or speed.
Flying approaches according to AoA  will keep the pilot workload to minimum, and removes possibilities of mistakes in calculations.
The pilot can simply monitor the AoA, and keep it in target value while maintaining glide path by adjusting the throttle. No math, tables or mnemonics necessary.

Some simple math is needed to compute a corresponding airspeed to cross check against AOA to ensure it is correct.

Also, stall AoA stays the same when in turning flight. If you fly some wild turning pattern to land after having dropped some payload but not all, determining stall speed beforehand for each bit of the turn will be much more difficult than just fly with the AoA scale I would imagine

While I don't disagree with the other answers, another reason is that AoA is more difficult to measure and is only useful for some parts of the flight (not useful in cruise, for example).  So, most GA aircraft, and many airliners do not have an AoA sensor or AoA display.  It generally involves adding a mechanical wind vane that sticks out into the airstream and is connected to an encoder.  This position then has to be displayed to the pilot.  It can't ice up, or be too sensitive to debris impact etc.  This all adds cost and complexity.  It also adds complexity to pilot operations as it involves an additional display for the pilot scan.
Of course, it can be done, but many aircraft instead only display airspeed.  Even pitch is not required for VFR operations, but is for IFR.  Flight instructors often repeat, "Look outside!" for pitch.
Meanwhile, airspeed measurement for aircraft has been around since the Wright Bros

I would add that newer designs for AOA sensors have mitigated these issues somewhat. The system I have on my plane (from Dynon) involves a pitot probe with two orifices, one oriented normal to the flight path that measures dynamic pressure to determine indicated airspeed, and another oriented 45 degrees down. a small computer is used to determine AOA by comparing the difference in pressure between these two orifices. No moving parts are involved, and the icing issue is resolved as the pitot probe is (can be) heated just like any pitot probe.

As to the comment about additional pilot work load, although not in common use, there are many options for communicating AOA to the pilot, some of which actually reduce pilot workload.

Since maintaining AOA at that  AOA where your aircraft is safe to fly at on approach is guaranteed to put you at the optimum approach airspeed considering all factors (gross weight, configuration etc.), one option (used in many military aircraft), presents AOA as a generated aural tone which starts out as a low prf pulsing tone at somewhere around L/D max AOA, increases in prf until just below optimum approach AOA, where it changes to a steady tone at optimum AOA, & changes to a high prf high freq stall warning  tone near stall. With this system pilots need not look inside the cockpit at all.

Most if not all airliners younger than a Boeing 767 should have AoA vanes though.

We may anticipate that most fighter planes are not staticly stable.
When an aircraft is staticly stable, airspeed and angle of attack are set by elevator trim.  While on a glide path, the aircraft will self adjust it airspeed and maintain its trimmed AoA.
For an aircraft with nuetral (or even negative) static stability, flying a glide path will involve careful monitoring of airspeed and angle of attack.
Flying by airspeed would involve knowing the aircraft weight, allowing a "safe" amount of elevator trim.
Flying by angle of attack allows the plane to drop into its "natural" linear glide path.  From here power is added or subtracted to reach the ground aiming point (generally a runway).
In reality "flying by airspeed" (how your scarf ruffles) and flying by AoA go hand in hand.  For a given weight, this can actually be established in flight before approaching to land.
The AoA parameter gives the flight computer a level of redundancy to "fly this airspeed at that AoA".  But for a staticly unstable fighter, constant pitch adjustments will be required to maintain proper airspeed.

airspeed and angle of attack govern airfoil aerodynamics

Landing speed will be variable (based on weight), there for landing distance will be variable.
A case for longer runways, or good strong arrestor cables.


text:
Deriving Accurate Exit air velocity of rotor
My goal is to derive the equation for exit air velocity as a function of:

Pitch angle
RPM
Rotor Radius
Constants like ρ

Assumptions:

Vc is relative velocity of air upstream, and is = 0
Inflow is unevenly distributed in the form of tip loss
Out-of-plane velocity Up is much smaller than the in-plane velocity Ut so that U=Ut
vi=vh (hover)
Each rotor blade is a traditional linearly tapering design
Shrouds of any kind are not used to direct airflow

Definition of variables:

Pitch angle: theta
Density of the fluid (air in this case): ρ
induced inflow velocity in the plane of the rotor: vi
Rotor Thrust: T
Area of the rotor disk: A=π(R^2)
Number of blades: Nb
Coefficient of thrust: CT

Current work that I have done:
Used a combination of equations 2.58, 3.31, and 3.36 to derive an equation for vh as follows:

Next steps: derived torque equation (have some from chapters 2 and 3 but they are not a function of Nb)
Or just a more simple equation that outputs the output velocity as a function of rotor pitch, RPM, Diameter, and blade count
References:
Principles of Helicopter Aerodynamics by J. Gordon Leishman
Ultimate goal: derive rotor wash / prop blast as a function of:

vertical distance from rotor
horizontal distance from center of rotor

Is there anything I don't know enough to ask? feel free to educate me from the ground up, I have only read 3 chapters from a textbook, so I have a lot to learn.

I think that we cannot give a better explanation than the Leishman himself. Anyway if you need to have the blade's geometry in your equations then you need to use the blade element momentum theory and not the simpler momentum theory that you've been using so far.

Are there any more books about rotor / prop design that I should read? I've tried to use blade element momentum theory, but have trouble with calculating L (lift force) due to lack of data about the rotors since I only have pitch angle, diameter, blade count, and rpm, thanks!

Is this an university assignment? What's your background? What are the precise known inputs and requested outputs?

Its for a project involving a 1x1ft square of metal that is being hit with the prop blast of a variable drone (variable diameter, motor number, rotor diameter, rotor blade number, and rotor blade pitch, motor rpm, and height from 1x1ft square) assuming the center of the drone is also on the center of the square. The goal is to simulate the amount of force or "weight" put on the 1x1ft square from the air produced by the motion of the props. Assuming hover conditions.

The GUI I did for it so you can get a visual feel for it: https://imgur.com/a/a6DZ4DZ


text:
How to command autopilot to start a fixed rate descent?
Is it possible to command an autopilot to start a fixed rate descent? For example to avoid weather during the cruise phase? From what I have read the autopilot would maintain constant speed and altitude until interrupted but havent found a fixed rate descent function.

The answer greatly depends on the type of autopilot used. Small cessna's and the like generally have less advanced autoflight systems than aircraft made by Airbus or Boeing.

Here is what typical autopilot controls look like

Image source: What is a "Flight Director"?
The pilot can set the required altitude, as well as the required vertical speed to get to that altitude. A cruise decent will typically use a shallower descent than an approach descent.

This answer explains that there is a flight level change mode and a vertical speed mode. The latter will hold vertical speed until a target altitude is reached.


text:
What is the minimum power system required to fly a commercial aircraft?
I heard of a A320 landing, no lights, both generators out, RAT deployed, and with only 28VDC(battery power),  made a successful landing. Is this possible?

If the RAT deployed, the A320 would have lost both generators and the APU

I heard of a A320 landing, no lights, both generators out, RAT deployed ... , made a successful landing. Is this possible?

Yes

A United Airlines Airbus A320-200, registration N409UA performing flight UA-497 from New Orleans,LA to San Francisco,CA (USA) with 109 people on board, was in the initial climb when the crew reported smoke in the cockpit, levelled off at 5000 feet and returned to New Orleans. The crew reported before joining downwind that they had lost all instruments and requested to be talked down by ATC via Precision Approach Radar (PAR). The crew descended to 600 feet where they got visual contact with the water of Lake Pontchartrain and continued visually for a landing on runway 19 about 10 minutes after departure. During landing the aircraft blew both right hand main gear tyres, went left off the runway, stopped with all gear just off the paved surface north of the intersection with runway 10/28, and was evacuated via slides.
A number of passengers needed medical attention due to smoke inhalation.
Post landing photos showed the RAM Air Turbine (RAT) deployed. Runway 01/19 was closed for about 10 hours.
The crew told passengers that they had lost all electronics and were flying on minimal backup systems, landing would occur overweight with minimal braking and minimal steering ability.

From http://avherald.com/h?article=43a6bc08

Early on, NTSB investigators suspected a faulty fire-warning sensor in the cargo hold. And they tentatively concluded that the cockpit crew had skipped a portion of the pre-flight checklist then failed to restore power to some equipment, making it more difficult to carry out the emergency landing

From http://www.nola.com/traffic/index.ssf/2013/05/united_497_emergency_landing_i.html
Basically it seems there was nothing wrong with the aircraft apart from a faulty smoke sensor. The crew missed this in their preflight checks, noticed it after take-off and then turned off all the electric generators (they were supposed to turn one back on according to the checklist - they didn't). Anyway, they landed the plane OK on emergency power from batteries and RAT.

with only 28VDC(battery power)*

On the A320 the RAT supplies the ESS bus, not main buses 1 and 2. It powers both the AC ESS bus and the DC ESS bus through the ESS TR (transformer rectifier). So that's not just 28V DC but AC too.
In the interval after both main generators failing and APU being unavailable, I believe the DC batteries power the DC ESS bus until the RAT is deployed and providing power (which should happen automatically and quickly). So in this short interval there is only DC power on the DC ESS bus.

Thank you for your answer-I wasn't aware of the UA-497, although I had heard of a landing at Juneau, Alaska by an A320 with only battery power. I failed to mention the APU was disabled, as this would have provided all essential power, and the crew took off with #1 Gen. out, which was allowed given #2 and the APU were both functional.


text:
What is the ground idle fuel flow of an A320 (CFM or IAE or neo)?
Standing on ground, what is the idle fuel flow of an A320? ISA conditions, both engines running?

I don’t normally post answers that I cannot personally verify, but I have a Facebook friend who is FO in an A320. I sent him a message and asked your question. I don’t have any official documentation to reference, so take this as anecdotal...
An A320 with CFM engines burns about 600lb/hr per engine at idle in ISA conditions.
An A320 with V2500 engines burns about 730lb/hr per engine at idle in ISA conditions.
If the APU is running, it will burn another 300lb/hr.

I guess that is with the APU running the AC packs?

@JohnK: That would be my guess. He is based out of Texas, so...


text:
How did War Emergency Power work?
Someone can explain how WEP (War Emergency Power) worked on the planes of the Second World War? And if it was really effective during a combat?

is it related to water injection?

@ManuH Not directly.  Water injection was sometimes used to reduce detonation at very high manifold pressures, but War Emergency Power could be used without it -- water injection was commonly used for Takeoff Power to avoid the stress on the engine that War Emergency Power produced.

Generally, War Emergency Power was a power setting that was available, but exceeded the engine's long term running condition rating.  It was usually specified in manifold pressure, and primarily applied to supercharged or turbocharged engines (which was most of the high performance piston engines by the end of the War).
The way it works was much like how it works in modern jets (where it often specifies afterburner operation): There was a stop on the throttle and a warning mark on the manifold pressure gauge that warned the pilot not to exceed those settings, as well as a specification in the manual for the longest permitted time in War Emergency Power setting.  To engage this setting, the pilot either had to manipulate the throttle lever in an unusual direction (pull up, push sideways) or break an actual wire with a push strong enough not to be done by accident, in order to bypass the stop.  In aircraft where it was used, the wire served to notify ground crews that WEP had been used.
Once engaged, War Emergency Power would give anything from 10% to 20% more total horsepower (in most cases, though the last version of the P-51 got roundly 60% increase), and a significantly increased rate of engine parts wear and risk of failure (blown gaskets, thrown connecting rods, even holed pistons from detonation if the fuel wasn't quite up to spec).  This doesn't seem like much, but 20% additional power would amount to 10 knots or more of additional level flight airspeed, or several hundred feet per minute of increase in climb rate -- it could make all the difference in air combat, when fighters of one force were often quite closely matched to those of the other in terms of performance.

In other words, it was just like "push to pass" is in modern auto racing. That gives you an increase in turbo boost or battery power so you can make a pass on the car in front of you (or defend against the car behind), but it causes additional engine wear or uses battery power that takes time to regenerate. It's a short-term boost that can come at a long-term cost.

@FreeMan Pretty much.  The manufacturer would rate the engine for so many minutes in War Emergency Power and require a detail inspection afterward -- minimum of an oil change (for inline engines that didn't run total-loss oiling), stick a borescope in every spark plug hole to check condition of pistons, etc.  Much of this didn't get done at a front line base, of course, but there was a lot of "built way better than claimed" in those days, at least in US aircraft.

The limit was normally set by copper witness wire across the throttle slot which set the normal wide open throttle position, called the "gate".  Copper wire was used to provide a clear indication to maintenance that WEP was selected in the event that the pilot neglected to bring it up. Copper witness wire provided a usable mechanical stop for setting the normal forward limit of throttle travel, but could be broken by a sufficiently strong push. Going to WEP was "breaking the gates" or "busting the gates" or "through the gates".  My father flew C-47s in WW2 and used WEP once to clear terrain.


text:
Why don't modern F-16s incorporate the VISTA thrust vectoring?
Modern incarnations of the F-16, such as the Block 70/72, F-16I and F-16IQ appear to incorporate many modern technologies. These aircraft are also rather different from the original F-16A, being heavier and more of an all-round fighter than a dedicated air superiority aircraft.
With heavy wing loading and a heavier aircraft overall, why was it considered to drop incorporating thrust vectoring into the F-16, to retain more of its dogfighting capability? This is hardly new technology to the F-16, considering the VISTA first flew in 1992.
This appears especially interesting considering highly maneuverable opponents such as the emerging J-20, Su-57; but also older fourth generation aircraft. The Typhoon, Rafale, Su-37, J-10, and others, are all roughly in the same ballpark as the more modern (heavier) F-16s, correct me if I'm wrong.
Does this indicate that dogfighting is a thing of the past? How can this be, considering that the F-22 was designed to be supermaneuverable? This appears to be a contradiction to me.

Thrust vectoring adds weight, cost and complexity, and the benefit isn't important enough to outweigh that.
Various aircraft have been used to experiment with thrust vectoring, but the list of thrust vectoring fighters in service is fairly short:

F-22
Su-30 (newer variants)
Su-35S
J-20 (fairly new)
Su-57 (very new)

The F-16 was envisioned to be a small, light, and cheap fighter. It has certainly grown heavier and more expensive over the years, but that is still one of the big selling points. Although it worked fine for the VISTA experiments, the limited space, power and cooling on the F-16 is best used for other upgrades, such as better sensors. And the F-16 is still selling fairly well. Lockheed is still getting enough orders that they moved the production line to make room for the F-35 rather than shutting it down.
Even the F-35A, intended to replace the F-16, adds stealth but not thrust vectoring. If you can kill your adversary before they even know you're there, dogfighting ability becomes less important. There's also the reality that US manufacturers build primarily for the US, and for US interests. There might be resistance to exporting an even more maneuverable F-16 when it may already be close in capability to the F-35.

Could it also be that modern FBW controls make the F-16 sufficiently agile without thrust vectoring? If its turning ability is limited by engine power and g-limits, then thrust vectoring wouldn't add anything.

Also, a testbed isn't automatically a product. The list of thrust-vectored aircraft contains a who-is-who of modern US aircraft, of which most are experimental. You want to test the applicability of 2D and 3D thrust vectoring, on one- and two-engined jets, and possibly on different configurations. Yet there is a long way from testing to product.


text:
What is the average deviation from the PAPI glide slope at the threshold?
If one approaches with three red and one white light what will be the average deviation from the standard threshold height of 50 feet?

Most PAPI installations are set up for 3 degrees with 2 red and 2 white. 3 red and 1 white would be about 2.5 degrees and if you are handy with math, you could figure out the threshold crossing height.

If one approaches with three red and one white light what will be the
  average deviation from the standard threshold height of 50 feet?

Short answer
It depends on the location of the PAPI (and its slope). For a 3° PAPI at 1,000 ft (305 m) from the threshold, three red lights means you cross the threshold between 43 and 49 ft.

PAPI heights at runway threshold

Explanation
There are different ways to calculate this. One is to start from the distance of the change from  red to white for the third light, which is given in this table (line B):

Source
The right side is for APAPI, let's ignore it. d2 is the distance from the PAPI (not from the threshold) an aircraft flying at a height of 1,000 ft sees the system switching from 3 red lights to 2 red lights.
If at this point the aircraft starts following the B glideslope, it will maintain this 2/3 red lights limit until landed. It will have lost 1,000 ft over a distance of d2.
A simple application of Thales theorem, about proportionality, says:
height at threshold = 1,000 * d / d2

where d = distance PAPI to threshold.
Example:

PAPI slope 3°
d2 = 3.33 NM = 6,167 m
Distance PAPI to threshold = 305 m

so height at threshold = 1,000 * 305 / 6,167 = 49 ft
If we do the same calculation for the other light transitions, we get: 60, 55, 49, 43 ft respectively.
Actual height
To elaborate on CVn's comment: A PAPI isn't corrected for parallax error. The sensor being the eyes, and having a difference between eyes height and wheels height, the PAPI glide path is always higher than the landing wheels.
In particular when the glide path crosses the threshold at 50 ft, the wheels are a bit lower, the actual height depending on the particular aircraft configuration and attitude. This error comes in addition of the glide path inaccuracy itself (+/- 3 ft at the threshold between lights + angular setup error). A PAPI is not a precision instrument, and the landing has to be done visually.
Larger aircraft performing precision landings are fitted with radar-altimeters located on the belly, well below the pilot eyes and close to the main wheel. Their raw value is usually corrected, by adjusting the aircraft installation delay beyond what is required for line length cancellation, to indicate the main wheels height.
This improved precision allows for lower decision heights (the point where the pilot is not allowed to pursue the landing if the runway isn't in sight).

Might be worth noting that if you're flying the PAPI, your eyes are at the glidepath. Even for small airplanes, this will probably be on the order of 1.5-2 meters above the wheels; more once you pitch up. For large airliners, it will be a significant difference.


text:
Why are heading in PFD and map off by exactly 10 degrees?
I was flying on my X-plane 11 flight simulator today and noticed something quite unsettling.
This was a VFR flight between KIAD and KDCA. I pulled up my map for reference and noticed I was flying orthogonal to the map's north direction, even though I was flying a heading of 100.

The image shows the map and the PFD for your reference.
What is going on? Did I mess with some heading calibration setup? 100 degrees and 90 degrees are perfectly 10 degrees off, so I don't think this has to do with magnetic north issues.
P.S. Strangely, I got to my destination simply following the headings from ATC.

Winds aloft possibly? One is ground track, the other is heading, they can often be different with wind...

related (not sure this is a dupe, your question being more specific): What are the differences between Bearing vs Course vs Direction vs Heading vs Track?

another related question: How to calculations deviation on a magnetic compass and effect of it?

KIAD has a magnetic variation of 10 degrees west, and KDCA has a magnetic variation of 11 degrees west, so true vs. magnetic north actually makes sense.

I'm a little new to navigation.. I could benefit from it if you explain a little more, or redirect me to some answers which address this. I tried a search or two, but couldn't find much.

@WilliamR.Ebenezer I have closed your question as a duplicate of another which will fully explain the difference between true and magnetic north.

@Jamiec Thanks.


text:
Why is parking brake set OFF before jacking?
I have a question about the procedure to jack the aircraft for wheel replacement. According to the manual, it required to set the Parking brake OFF before starting jacking at the Main Landing Gear.
I curiously wonder if the brake is set OFF to let the wheels on the other landing gear could rotate freely on its axle.
Why don't we keep the parking brake ON to prevent the aircraft moving and slip out of the jack?

Thanks for your comment. I have read that question. However, my question is about setting braking OFF before jacking aircraft which happens in a short period of time.

Probably makes things easier to get the old wheel off & the new wheel on.

Absolutely NOT a duplicate.

@dingdingdong Oh yes, of course. Absolutely my mistake. I must have completely misread "jacking" as "chocking" or something like that.

Does this answer your question? Why are parking brakes not used for parking jetliners?

@KorvinStarmast Unfortunately, it isn’t the answer for my question.  That question is about using parking brake when planes park for long period of time, whereas mine is for Jacking for wheel replacement.

The wheel is removed with the PB depressurized so that the discs are free to float (both rotationally and slightly axially) when the wheel is being slid off, to prevent any potential binding of the disc projections where they engage the slots in the bore of the wheel as the wheel is pulled out. If the discs are locked in place by brake engagement, any off-axis movement of the wheel unit as it's being slid off could jam it on the discs, making it really hard to pull off.
When the procedure is started, the airplane is chocked at the other gears, so the parking brake is already made redundant at this point.  As a former aviation technical writer in another life (not Airbus), I would say it was done that way for work flow purposes (getting all the cockpit related precautions out of the way before you start the procedure) because you don't need the PB on once chocked, and you're going to have to release it anyway, so it's done during the preparation steps prior to the actual jacking.


text:
Why does the APU Master Switch flash during a normal APU shutdown?
Hi all and Happy New Year!!!
I have a question relating to APU shutdown procedures on Airbus A320. During normal APU shutdown sequences, I noticed that after around 1m20s, the FAULT legend on APU master switch will flash.
I wonder if the FAULT legend comes on to indicate APU intake flap is closed or any other reasons.


text:
How to calculate power required to drive an isolated fan of a TurboFan engine?
If I isolate the fan from a TurboFan jet engine (specifically GE90) and power it electrically with a motor,

how much power is required for the electrical source to drive the fan to give the same amount of thrust as a regular GE90?
in a regular GE90, what is the thrust produced by the fan alone (excluding the thrust from the combustion of fuel)?

https://en.wikipedia.org/wiki/General_Electric_GE90:

The LM9000 is an aeroderivative gas turbine available in two options; the LM9000 without water augmentation outputting 66 MW (89,000 hp) at a 42.4% efficiency before cogeneration, and the LM9000 with water augmentation outputting 75 MW (101,000 hp) at a 42.7% efficiency before cogeneration.

An aeroderivative gas turbine is basically the same turbine with a driveshaft connected to the low pressure stage instead of the fan. So its shaft output power would be roughly equivalent to the power that goes to the fan in the turbofan version.
The 66MW (the aero version does not use water augmentation) figure corresponds to max continuous power or a bit less. The take-off power is even more.

1) Based on flow rate and nozzle exit velocity you can calculate the minimal power for acceleration of the air. This assumes 100% efficiency of the fan and nozzle. Actual power will be more because they are not 100% efficient.
2) Estimate the power output of the low pressure turbine, which roughly equals to the power input to the fan. This could be done by connecting the LP turbine to a dyno and operate the assemble as a turbo shaft engine, or roughly estimate it by fuel consumption rate, fuel energy density, and thermal efficiency.
3) Model the takeoff process as a constant-power process, calculate the accumulated kinetic energy of the airplane and divide it by time and number of engines, which is a even more ideal case than 1). This ignores the most inefficiencies and should give the lowest number.
Remember, power isn't a concept tied to electricity. So the power to drive the fan has noting to do whether a electric motor is driving it or the LP turbine is driving it. The numbers are the same (if you live in a region where mechanical power and electric power are measured in the same units).
Edit:
Thinking about it again, I think this question got the logic wrong.
Before the design work of a turbo fan engine, one of the most basic spec is how much power do you want to drive it with (other questions include bypass ratio and thrust).
This is not something you estimate after you have an actual engine. It's something you have to know before the engine is designed (and simulated when it's designed, and tested after it's designed).
The quickest way to know the power required to drive GE90's fan is to ask GE engineers.


text:
Is drag coefficient lowest at zero angle of attack?
The drag coefficient of a symmetric airfoil is lowest when its angle of attack is zero. I'm not sure if this is true in general.

Welcome to Av.SE!

Obviously, drag should be smallest for symmetrical airfoils at zero angle of attack.
However, most airfoils have camber, and then the lowest drag is at positive lift coefficients in case of positive camber. Where that point is exactly depends on many parameters; in case of laminar airfoils even local imperfections can have a noticeable effect. Generally, the lowest drag can be found at the angle of attack where the stagnation point is exactly at the center of the leading edge, where the local curvature is highest. A deviation from this point will force the flow on one side to negotiate this point of highest curvature all by itself, resulting in a suction peak which will increase the losses in the boundary layer.

This is a theoretical drag polar (calculated with XFOIL) of an airfoil with a 20% camber flap at different settings and a Reynolds number of 1.5 million. The laminar bucket is clearly visible and produces a range of lift coefficients with nearly identical drag. The small waviness at the lower end of the laminar bucket is an artificial result of smoothing the plot.
What is obvious is how camber shifts the area of minimum drag up and down. If you use the right side of the plot to find the angle of attack of minimum drag, you will find that this is not constant but goes down as flap angles go up. For the 0° flap polar it is at about -2° AoA. This is caused by the induced angle of attack which increases with the lift coefficient.
The 6-series NACA airfoils were the first to be systematically designed with the pressure distribution in mind, and the design lift coefficient is where the condition of the ideal stagnation point location is met. This is indicated by the digit right after the hyphen in the airfoil name: Divide this digit by 10 and you have the lift coefficient of minimum drag.
Example: The $63_1-412$ airfoil has its lowest drag at a lift coefficient of 0.4.
If you want to know the angle of attack with the lowest drag of a whole airplane, this is a very different matter and needs to include the drag due to lift, which is of course smallest at the zero lift polar point.


text:
What are the effects of dihedral on net lift of a wing?
I read somewhere on the internet that dihedral decreases wing lift,so my question is will dihedral decrease the lift of an airplane wing since the wings are attached to the fuselage  at an angle ? If yes,will anhedral also decrease the lift produced by the wings?
If dihedral/anhedral decrease wing lift,is the decrease in lift negligible im comparison to thesame wing(with thesame area,aspect ratio) but without dihedral?

This is answered here but the title of the question does not help locate it.

The effect of dihedral on lift can be obtained applying high school trigonometry:
$$
L=L_0 \, cos \,\theta
$$
where $L_0$ is the lift of the original flat wing and $\theta$ is the dihedral angle. This holds true for negative values of $\theta$ as well.
Since the cosine can be approximated by unity for small angles, the net effect is negligible in most practical cases, e.g. $ cos(5^o)= 0.996$


text:
Where can PT-6 engine series starter-Generator diagrams be found?
Wondering if anyone can direct me to the spec sheet, diagrams, and breakdowns of a typical Starter-Generator for a PT-6, say a PT6-66 series or similar. And also who are the major manufacturers of such units for turbo-props? Thank you.

Have you asked Pratt & Whitney for this information?

Do you have a link or a contact list for anyone in Pratt & Whitney who would supply this information to an outsider? And why would they do that...?!

I don't have a contact but I did provide a link to the P&W website in my comment above. You haven't said why you want the information. If you don't think P&W will help you then I doubt anyone else will. It may be covered by non-disclosure agreements, and almost certainly by copyright.


text:
Can the Landing gear be lowered when system A hydraulic pressure is lost in Boeing 737?
In the FCOM OF BOEING 737 HYDRAULIC SECTION I went through the part Landing gear transfer unit that it is used to raise the landing gear at a normal rate when system A EDP volume is lost.
My query is if only the hydraulic system A is lost with landing gear up position can it lowered using system B hydraulic or do we need to use manual gravity gear extension?

Related: this answer of the question "If the landing gear is stuck in an airliner, can the crew go down and manually open it?". It speaks about gravity extension, but nothing about the hydraulic of the B737.

Normal gear extension is by the A system; alternate gear extension is by manually releasing the up-locks, allowing the gear to free-fall into place. There is no provision for any other alternate gear extension beyond that, such as via the B or Standby hydraulic systems, electrics, pneumatics, or otherwise.


text:
What is the purposes of grid-like pattern route for small aircraft?
In these days, I see small aircraft like P68 fly with a very accurate "grid-like" pattern. I'm sorry but I don't know the correct word. Here a couple of examples, grabbed today at the same time:

and

What kind of job they might do with these routes?
I don't think it's about agriculture because the patterns are too long and they fly over cities and sea as well (and I don't think this kind of aircraft is the most suitable for such a task).
I thought about observation of restriction for COVID-19, but I would use helicopters instead of aircraft, and again the patterns are so long for just be sure if anyone goes out for a walk.
Any sort of collecting photographic data? Like Google does for its Maps?

Related: Why did this aircraft fly in this pattern over Southern Italy?

Welcome to AviationStackExchange. It seems that this question has popped up at lest once a month from newbie posters from the same region. This may or may not have anything to do with the current pandemic. This may be a scheduled aerial survey that goes unnoticed when people have more important activities taking their attention.

@DeanF. yes, it definitely makes sense. During "normal" days there are so many aircraft and it's very hard to spot this particular pattern. Now there are very few flights so it's very easy to spot them!

I tried to edit my comment to add info on the fleet operator for this and other posters’ questions. I made it an answer below instead.

The aircraft flown in the previous post from over three years ago is not listed in the fleet of aircraft for the company in question in your post. But, it was probably flying the same or similar mission.

These are most likely photography flights for cartography. Due to the COVID crisis, there are many flights like these lately because it is now easy to obtain permission from Air Traffic Control for these patterns.

I doubt this has anything to do with the lockdown. Lots of these flights take place every year, corona or not.

@J.Hougaard This flight could be unrelated to the lockdown. But there are many of these photo flights taking place now, much more than usually and especially in CTR's. This specific flight is operating in the CTR of Venice and probably also in the CTR of Treviso.

This may or may not have anything to do with the covid-19 pandemic. This may be a scheduled aerial survey that goes unnoticed when people have more important activities taking their attention.
As a matter of fact, the last couple of newbies posted flights from the same company as you have posted today, Compagnia Generale Ripreseaeree Partenavia . You have posted a different aircraft from their fleet. But, it is the same private company.

@ManuH - Other than removing the welcome, how else would it be edited without changing the answer? Also, the answer includes the aerial survey company which owns and operates this aircraft. The aircraft in the previously mentioned question marking this one as a duplicate is from a different company. Though, they are probably fulfilling the same or similar missions.


text:
When using the Source Panel method and dealing only with sources and sinks, why do lift and drag equal zero?
I am trying to deal with the airfoil with the "Source Panel method", and when I try to calculate the lift and drag, they always equal to zero. And I have found an explanation in some articles, that sources and sinks don’t create any drag/lift. But I don't understand the reason. Could someone give me some explanation?

Welcome to [aviation.se]! Your question is collecting close votes because it's not really clear what you're asking. Please [edit] you question to include more details. Perhaps include the formula(s) you're using and the numbers that you are plugging into it. Someone might spot some errors there and be able to guide you to correcting them. As it stands, you haven't really given anyone much of anything to go on. If you'll take the [tour] and read the [help], especially on "how to ask", you'll get a better idea of what to provide to get a good answer to your question.

@FreeMan Seems pretty clear to me what he's asking

You should add links to the articles you don't understand so that it is easier to point out unclear passages.

A single source, or a source sheet comprised of infinitesimal sources (sinks are just sources with negative strength), produces a velocity field that contains no circulation. Without circulation, it can't possibly generate any lift. The Source Panel Method is just a discretized computational method for source sheet. It, therefore, only produces non-lifting flow.
Drag, in the context of incompressible potential flow, only arises through induced drag. Without lift, induced drag would also be nil.
To predict lifting flow, vortex sheet modeling is typically used that, by construction, produces non-zero circulation (unless the integral of sheet strength is zero by prescription). An additional prescription of the Kutta condition is required to guarantee a unique physically correct lifting flow.

Image collated from: http://web.mit.edu/16.unified/www/SPRING/fluids/Lecture_Notes/f18.pdf
A mathematically equivalent methodology is to use doublet sheet modeling, using the following tranform:
$$\vec{\gamma}=\hat{n} \times \tilde{\nabla}\mu$$
where $\vec{\gamma}$ is the vortex sheet strength, $\hat{n}$ is the sheet normal, $\mu$ is the doublet strength, and $\tilde{\nabla}$ is the surface-gradient operator.
Somewhat confusingly, a doublet is a combination of a source and sink that are separated by an infinitely small distance. A single doublet is non-lifting. However, a lifting flow can be generated if the doublet sheet varies in strength. This video based on Drela, Flight Vehicle Aerodynamics provides an extremely good intuitive explanation.

It's great! Thanks for the explanation! It helped me a lot


text:
What are the differences between turbojet and turbofan (low and high by-pass ratio) in the graph of thrust relative to speed?
What are the curves of thrust relative to velocity of the turbojet, low-bypass turbofan and high-bypass turbofan engines from static thrust to Mach 4 generally. Approximation is okay as long it has any Mach or other speed values. Graph is preferred, because most graphs of these values I have found are starting at around Mach 0.5 or completely without any axes description.
Mostly I want to know is what difference there is at speeds of 0 to Mach 1 in these three engine types.
(I have wrote question containing around 400 words so I reduced it to this after getting suggestion its too long)

Fixed @RalphJ, I hope it is more clear now.

Much better, thanks!


text:
Does Vx & Vy calculations for aircraft with fixed pitch props account for lower power output at lower rpm?
In an aircraft with a fixed pitch prop, engine rpm can change significantly (increases) as airpeed increases. Since the power output of the engine increases as rpm increases, clearly the calculated Vx & Vy speeds should take this into consideration, especially on aircraft with a wide range of speeds between takeoff/climb and cruise speed, and have a "cruise" prop mounted (designed to reach maximum engine power/rpm at or near cruise airspeed). Do the published Vx & Vy speeds in aircraft POHs take this into consideration? And how should a pilot who has changed his/her prop modify a published Vx or Vy speed to reflect this?

"Do the published Vx & Vy speeds in aircraft POHs take this into consideration?" -- I should hope so!  It's hard to imagine why this would be doubted--

How can they know what engine, or what prop, is gping to be installed on the plane? In the experimental world, every plane is unique, owner/builders are free to install almost any engine/propeller combination they wish.

Yes, the different amount of power that can be developed at different airspeeds is taken into account. If it were not, then Vy would always be equal to the minimum sink speed when the engine is off. Changing the prop changes the shape of the power curve and hence changes both Vx and Vy.
With a new prop, you can simply test your climb performance at different airspeeds. With a cruise prop you would expect the airspeeds to be higher than for a climb prop.

As I surmised.  So for those of us with experimental aircraft, (canards in my case), where we get to choose what prop (and what engine) we install, we really need to perform flight testing to determine what Vx and Vy speeds to use.

@CharlesBretana You'd need that flight testing in any case. Every experimental aircraft is unique. Even with the same engine and prop, there's no guarantee the V speeds are exactly the same. Changing the prop and the engine just makes the difference larger. AC 90-89C suggests that determining Vx and Vy should be done during the 11th to 20th hour of flight testing during part of the overall evaluation of its properties in slow flight.

Yes, that makes sense. I did not build or flight test my aircraft, I purchased it from the builder, who (according to the FAA paperwork), did, so I was misled by the published Vx & Vy speeds in my aircrafts POH. As I had never thought about the effect of the prop on these speeds, so seeing them published there gave me erroneous impression that they were applicable to the aerodynamic characteristics of the airframe itself, and not dependent on the prop and engine.


text:
How to understand turbo machines?(compressors,fans,pumps,turbines,etc..)
From what I understood, the energy transfer between the blades and the fluid(fluid to blade or blade to fluid) occurs due to the curvature of the blade.Its because of this curvature the stream of the fluid is redirected, the whirl component of velocity changes, by applying momentum equation(newtons second law) a force acts on the fluid and then by newtons third law we can see that the blade is pushed in the opposite direction.But sometimes during lecture my professor used to say the blades have aerofoil cross section.But he never explained how that is significant in a compressor.Please explain.

Axial compressor blades are wings.  The airfoil cross section maximizes the efficiency of the compressor blade at generating lift to generate torque in the case of a turbine blade, or generating lift to create downwash that forces the air into a smaller space in the case of a compressor blade, compared to a blade that was flat or had a simple curve.
Pretty much same as a wing or propeller blade.  I can make a plane fly with sheets of plywood for wings, but it's way more efficient to use airfoil shapes in redirecting the air to generate the force I want.

So just a curve or slope(to redirect the stream) is sufficient for energy transfer.The only puropse of aerofoil cross section is to improve efficiency.Is this right?

Please notify me if this is not the right forum to ask this question.

@Mohan, yes, you could use flat plates for compressor blades and it would work.  It just would not be very efficient.  Efficiency is extremely important in jet engines, that's why designers spend so much time to get the exact perfect shape.

@Daniel Kiracofe Thank you.Nice answer, that clears my doubt.

Note that the main job of the stator vanes is to redirect and control the downwash of the upstream row of compressor blades so that the flow hits the next row of blades at the optimum angle for that row of blades.  That is why they are variable on pretty much all modern axial compressors and are continuously adjusted by the fuel control system.


text:
Why is cloud base estimation taught in Fahrenheit?
I've seen the formula for estimating cloud base consistently presented using Fahrenheit temperature and dewpoint (for example, see PHAK 12-14; Jeppesen also teaches it this way):
$$
\mathrm{cloud\:base} \approx \frac{(\mathrm{temperature\:°F})-(\mathrm{dewpoint\:°F})}{4.4\mathrm{°F}} \times 1000'
$$
However, a METAR always gives temperature and dewpoint in Celsius, and pretty much everything else in aviation uses Celsius — even lapse rate, which is the basis for this formula, is taught as "2°C per 1000 feet". It seems to me that using °F in this formula invites error, since temperature and dewpoint must first be converted to Fahrenheit before using this formula. If the constant on the bottom of the fraction were $2.4 \mathrm{°C}$ then the formula could be used directly with numbers from the METAR.
Is there a specific reason why this formula is given in °F?

Where are you doing your studying?  I ask because in the US, I can't instantly remember any temperatures in anything other than Fahrenheit...

I’m studying in the U.S. (Washington DC area). However, I think all AWOS systems I’ve interacted with give temps/dewpoints in Celsius.

Why is cloud base estimation taught in Fahrenheit?

It's not, the source you are looking at is taught in Fahrenheit.

Jeppesen being an American company, I would assume that their teaching material uses units of measurement common in America.
There are many places which have the same formula in Celsius such as the wikipedia entry on Cloud Base among others.
I suspect that pretty much everybody looking to calculate this would understand the difference between Fahrenheit and Celsius and adjust accordingly. That's not to say that errors are impossible, but errors are always possible when calculating anything.